Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2010 / B / 6 / i

Glue the two analytic functions on their overlap, where their values agree. Since $\alpha <\beta$, the two half-planes cover the whole complex plane. The gluing therefore defines an entire function. It is bounded outside a sufficiently large disk by the two given infinity bounds, and bounded on the remaining compact disk by continuity. The Liouville theorem then gives one common constant.

For the standard canonical Wiener-Hopf factorization, first assume the nonzero kernel has Index of a Wiener-Hopf kernel zero and a continuous analytic logarithm $\ell =\log F$ tending to zero at both ends of the strip, with sufficient decay for the Cauchy projections. Choose $\alpha <a<b<\beta$. Let ${\Gamma }_{+}$ be the line $\mathrm{Im}\zeta =a$ oriented left to right, with its factor used above that line; let ${\Gamma }_{-}$ be the line $\mathrm{Im}\zeta =b$ oriented left to right, with its factor used below that line.

For $a<\mathrm{Im}z<b$, closing the rectangle between the two lines and using the Cauchy integral formula givesThus $F=F_{+}/F_{-}$. Multiplying $F{w}_{+}=w_{-}$ gives $F_{+}{w}_{+}=F_{-}{w}_{-}$ in the overlap. The canonical factors are analytic, nonzero, and normalized at infinity in their respective half-planes, so part (a) yieldsContours may be shifted within the strip without changing their factors on the appropriate side. When necessary, a subtracted Cauchy kernel fixes convergence and the common normalization constant.

The printed hypotheses omit the index and factor-growth qualifications; $F\to 1$ alone does not ensure them. An explicit counterexample is $F(z)=(z+i)/(z-i)$ on $-1/2<\mathrm{Im}z<1/2$, with $w_{+}=(z-i)/(z+i)$ and $w_{-}=1$. Both $w$ functions are analytic and bounded at infinity in their prescribed half-planes, and the relation holds. If admissible analytic factors made their common product a nonzero constant, $w_{-}=1$ would force $F_{-}$ constant and hence $F_{+}$ proportional to $F$, impossible because $F$ has a pole at $i$ in the plus half-plane. The natural factors $F_{+}=z+i$, $F_{-}=z-i$ instead give the nonconstant common polynomial $z-i$. Thus the boxed constant conclusion is the intended index-zero result, not a consequence of the literal assumptions alone.